Metamath Proof Explorer
Description: Cancellation law for subtraction. (Contributed by NM, 8-Feb-2005)
|
|
Ref |
Expression |
|
Hypotheses |
negidi.1 |
⊢ 𝐴 ∈ ℂ |
|
|
pncan3i.2 |
⊢ 𝐵 ∈ ℂ |
|
|
subadd.3 |
⊢ 𝐶 ∈ ℂ |
|
Assertion |
subcani |
⊢ ( ( 𝐴 − 𝐵 ) = ( 𝐴 − 𝐶 ) ↔ 𝐵 = 𝐶 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
negidi.1 |
⊢ 𝐴 ∈ ℂ |
2 |
|
pncan3i.2 |
⊢ 𝐵 ∈ ℂ |
3 |
|
subadd.3 |
⊢ 𝐶 ∈ ℂ |
4 |
|
subcan |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) = ( 𝐴 − 𝐶 ) ↔ 𝐵 = 𝐶 ) ) |
5 |
1 2 3 4
|
mp3an |
⊢ ( ( 𝐴 − 𝐵 ) = ( 𝐴 − 𝐶 ) ↔ 𝐵 = 𝐶 ) |